Thought experiment - A machine that predicts equally likely events A thought experiment:
Hypothetically lets assume a machine can predict result of equally likely events correctly with probability 0.8(say a coin toss for example).
We know probability of getting a head/tail in a coin toss is 0.5, The machine beforehand predicts that head is the result of the next toss. Does this change the probability of getting a head/tail in the next toss?
(because we know the machine predicts 8/10 times correctly and also assume the machine is intelligent and once event has occurred the machine predicts the result with 100% accuracy)
Does the first statement make sense? Is it even possible to call the events equally likely if the machine can predict the outcome with some accuracy?
Bonus discussion:
Two cases I imagined: (Not sure if it is relevant to this forum)

*

*Either the machine can go into the future with some accuracy and return to the same timeline

*Or the machine knows every occurrence of tosses with some precision

Note: The machine cannot predict with 100% accuracy - This makes seeing the future with absolute precision not possible. (If not I think probability cannot exist right? we already know the future)
For me case 2 makes the event not equally likely because the machine always knows the future with some accuracy.
Case 1 - Not really sure because the event has to happen normally, its just that the machine knows the result by traveling to the future but this might have to do with closed time loop and paradoxes
Edited first statement from: Hypothetically lets assume a machine can predict result of equally likely events correctly 8/10 times(say a coin toss for example).
 A: Let $M_i\in \{H,T\}$ be the machine's prediction for coin toss $i$ and $X_i\in\{H,T\}$ be the actual outcome of coin toss $i$.
What we want to study is the joint distribution $P_{MX}(M=m,X=x)$:
$$P_{MX}(M=H,X=H) = P_X(X=H)P_M(M=H|X=H)=0.5*0.8 = 0.4 = P_{MX}(M=T,X=T)$$
By symmetry,
$$P_{MX}(M=H,X=T) = P_{MX}(M=T,X=H)=0.1$$
Therefore, the distribution of $(M,X)$ is given by:
$$P_{MX}(m,x) = 0.4 \mathbb{I}_{m=x} + 0.1\mathbb{I}_{m \neq x}$$
Note that the marginal probability $P_X(X=x) =P_{MX}(H,x) + P_{MX}(T,x) = 0.4+0.1=0.5$ as expected.
Also, $P_M(M=x) =P_{MX}(x,H) + P_{MX}(x,T) = 0.4+0.1=0.5$
So what is $P(X=H|M=H)$?:
$$P_{X|M}(X=H|M=H)=\frac{P_{XM}(X=H,M=H)}{P_M(H)} = \frac{0.4}{0.5} = 0.8 = P_{X|M}(X=T|M=T)$$
Again, not super surprising.
So you are much better off listening to the machine.
A: Provided the coin is fair, is it still possible to have a mechanism that predicts the outcomes correctly with some probability $>0.5$? Let $M_H$ denote event of machine predicting $H$ (heads).
Assume $P(M_H \mid H) = p$ for some $0< p\leqslant 1$. In your example $p=0.8$. Compute $P(H \mid M_H)$. By Bayes identity
$$ P(H\mid M_H) P(M_H) = P(M_H \mid H) P(H) = 0.5p $$
By total probability
$$P(M_H) = P(M_H\mid H)P(H) + P(M_H\mid T)P(T) = 0.5 $$
But then $P(H\mid M_H) = p$. This either contradicts fairness of the coin or $p=0.5$ is forced. Alternatively, the coin tossing could be Dependent on the machine, but that would defeat the purpose of "prediction".
But if $p=0.5$ is the best you can do, there's no point trying to build such a machine in the first place.

Notice what happens with $P(M_H)$. It only depends on the assumption that the coin is fair. So that would immediately tell you that in a fair game, there's no reason to have a bias toward any outcome. (duh!)
